A chemist dissolves 581.1mg of pure perchloric acid in enough water to make up 150 ml of solution. Calculate the pH of the solution. Round your answer to significant decimal places.

The pH of the solution is 1.41

Explanation:

To calculate the molarity of solution, we use the equation:

Given mass of perchloric acid = 581.1 mg = 0.5811 g   (Conversion factor:  1 g = 1000 mg)

Molar mass of perchloric acid = 100.5 g/mol

Volume of solution = 150 mL

Putting values in above equation, we get:

1 mole of perchloric acid produces 1 mole of ions and 1 mole of ions

To calculate the pH of the solution, we use the equation:

We are given:

Putting values in above equation, we get:

Hence, the pH of the solution is 1.41

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: